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(D) The energy of a photon is given by $E = \frac{hc}{\lambda}$.Given $\lambda = 2414 \ \mathring{A} = 2414 	imes 10^{-10} \ m$,$h = 6.626 	imes 10^

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$497.7 \ kJ \ mol^{-1}$

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(D) The energy of a photon is given by $E = \frac{hc}{\lambda}$.Given $\lambda = 2414 \ \mathring{A} = 2414 	imes 10^{-10} \ m$,$h = 6.626 	imes 10^{-34} \ J \ s$,and $c = 3 	imes 10^8 \ m \ s^{-1}$.$E = \frac{6.626 	imes 10^{-34} 	imes 3 	imes 10^8}{2414 	imes 10^{-10}} \ J = 8.232 	imes 10^{-19} \ J$.This is the energy required to ionise one atom.For one mole of atoms,the ionisation energy is $E_{mole} = E 	imes N_A$,where $N_A = 6.022 	imes 10^{23} \ mol^{-1}$.$E_{mole} = 8.232 	imes 10^{-19} 	imes 6.022 	imes 10^{23} \ J \ mol^{-1} \approx 495751 \ J \ mol^{-1} \approx 495.75 \ kJ \ mol^{-1}$.Rounding to the nearest option,the value is $497.7 \ kJ \ mol^{-1}$.

The maximum wavelength of radiation that can ionise a sodium atom is $2414 \ \mathring{A}$. The ionisation energy of sodium per mole shall be

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